Imagine two protons moving in the opposite directions with a velocity $v$ or $\beta c$, where $\beta = v/c$. At $t = 0$, they are a distance $r$ apart one along the same line (see the picture). I am having troubles deriving the exact relation between electric and magnetic forces (or fields) in this case.

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It was pretty straight forward to derive this relation in case of protons moving along. The example of the derivation can be found in the answer to the following post:

Relativistic electromagnetism and electromagnetic forces on 2 protons.

The conversion factor is $F_{magnetic} = \beta^2 F_{electric}$ The only difference is that in my case I use total force transformation explicitly ($F_{\text{in stationary frame}} = F_{\text{in moving frame}} / \gamma $) instead transformation of fields as given in the answer.

Now, when we move to the problem of oppositely moving charges, the direction of magnetic force is reversed and it will be repulsive, as by reversing the direction of one of the protons, we reverse the direction of created by it magnetic field.

My solution is then the following:

1) We move to the rest frame of the (lets say) bottom proton, which we refer to as first. In its own frame it feels no magnetic force as it is stationary. Therefore, the force on it $F^{'}_{tot} = F^{'}_{el} = e E^{'}$ (where E^{'} is the field created by the upper proton).

To find that field we first need to find the velocity of the 2nd (upper) proton in rest frame of the 1st. From relativistic velocity addition we get:

$$ v_{e_2}^{'} = \frac{-v - v}{1 - \frac{-vv}{c^2}} = \frac{-2v}{1 +\frac{v^2}{c^2}}. $$

That leads to:

$$ \gamma^{'} = \frac{1}{\sqrt{1 - v_{e_2}^{'2}/c^2}} = \frac{1}{\sqrt{1 - \frac{4v^2}{(1+v^2/c^2)^2c^2}}} = \frac{1}{\sqrt{1 - \frac{4\beta^2}{(1+\beta^2)^2}}}.$$

Now lets find the field acting on the first proton from the second one in the first proton's frame of reference. In the rest frame of the second proton, the field is simply $E_{rest} = k e / r^2$. Since it is moving in the frame of the 1st proton, the 1st proton will feel a perpendicular field component of: $ E^{'} = \gamma^{'}E_{rest} $. Then: $$F^{'}_{tot} = F^{'}_{el} = e\gamma^{'}E_{rest}.$$

2) In the lab frame (where both protons move with velocity $v$ in opposite directions) the total force on the first proton will be $$F_{tot_{lab}} = F_{tot}{'}/ \gamma = e\frac{\gamma^{'}}{\gamma}E_{rest} , \,\, \text{where} \,\, \gamma = \frac{1}{\sqrt{1-\beta^2}}.$$

where electric force acting on the first proton from the second one will be $$ F_{el_{lab}} = e E_{lab} = e \gamma E_{rest}$$.

3) Since we know that both magnetic (from considerations in the beginning) and electric forces will be repulsive, then:

$$F_{mag_{lab}} = F_{tot_{lab}} - F_{el_{lab}} = e\frac{\gamma^{'}}{\gamma}E_{rest} - e \gamma E_{rest} = F_{el_{lab}} \left( \frac{ \gamma^{'} }{\gamma^2} -1 \right).$$

4) From result of two protons moving in the same direction, we know that factor should be $\beta^2$, therefore, it should be that:

$$ \left( \frac{ \gamma^{'} }{\gamma^2} -1 \right) = \frac{1-\beta^2}{\sqrt{1 - \frac{4\beta^2}{(1+\beta^2)^2}}} -1 == \beta^2 $$

When trying to reduce right hand side I arrive at $-\sqrt{\beta^2 + 1}$ instead of $\beta^2$. I even tried to use the help of SymPy expression simplification functions, but same result. Have I made a mistake in the reasoning or in simlification of the final expression?

P.S. Here the conversion factor between the forces is derived by explicitly assuming magnetic field form.. But I am interested how it can be derived using relativity.

EDIT: Was just an annoying mistake in the calculations. everything works.

  • $\begingroup$ I'm not sure I understand, which equation gives you $-\sqrt{\beta^2+1}$? The last step that you've written seems fine, $$\frac{1-\beta^2}{\sqrt{1 - 4\beta^2/(1+\beta^2)^2}} -1 =\beta^2,$$ as you require... Also, my answer to this related question might be useful: Conceptual question about special relativity in electrodynamics. $\endgroup$
    – Philip
    Jan 14, 2021 at 20:43
  • $\begingroup$ could you please show this explicitely, I will accept it as an answer then. $\endgroup$
    – kek
    Jan 14, 2021 at 21:21
  • $\begingroup$ somehow I got it just now. $\endgroup$
    – kek
    Jan 14, 2021 at 21:24
  • $\begingroup$ Ah good. I was just writing that it's nearly too trivial to explain :) Good to know you got it! :) $\endgroup$
    – Philip
    Jan 14, 2021 at 21:24
  • $\begingroup$ Should I delete a post, or just let it be, or write derivation myself as edit or answer? (I am new here) $\endgroup$
    – kek
    Jan 14, 2021 at 21:26

1 Answer 1


For whoever may struggle with the same problem:

$$ \sqrt{1-\frac{4\beta^2}{(1+\beta^2)^2}} = \sqrt{\frac{(1+\beta^2)^2 - 4\beta^2}{(1+\beta^2)^2}} = \sqrt{\frac{(1-\beta^2)^2 }{(1+\beta^2)^2}} = \frac{1-\beta^2 }{1+\beta^2} $$

Substituting back into the expression:

$$ \frac{1-\beta^2}{\frac{1-\beta^2 }{1+\beta^2} } -1 = 1+\beta^2 -1 = \beta^2. $$ So

$$F_{mag_{lab}} = \beta^2 F_{el_{lab}} $$


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